The acceleration due to gravity on the surface of earth is given by, \[g=\frac{GM}{R^2}\]
The acceleration due to gravity on the surface of moon is given by, \[g_m=\frac{GM_m}{R_m^2}\]
According to the question, $M_m=\frac{M}{81}$ and $R_m=\frac{R}{4}$.
\[\therefore g_m=\frac{G(M/81)}{(R/4)^2}=\frac{G(M/81)}{R^2/16}\] \[=\frac{16}{81}\left(\frac{GM}{R^2}\right)=\frac{16}{81}g\]
Putting $g=9.8$ $\text{m/s}^2$, we get, \[g_m=\frac{16}{81}×9.8=1.936\;\;\text{m/sec}^2\]